$\overline{AC}$ is $4$ units long $\overline{BC}$ is $10$ units long $\overline{AB}$ is $2\sqrt{29}$ units long What is $\cos(\angle ABC)$ ? $A$ $C$ $B$ $4$ $10$ $2\sqrt{29}$
Solution: SOH CAH TOA os = djacent over ypotenuse adjacent $= \overline{BC} = 10$ hypotenuse $= \overline{AB} = 2\sqrt{29}$ $\cos(\angle ABC )=\frac{10}{2\sqrt{29}}$ $=\dfrac{5\sqrt{29} }{29}$